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Question-208359

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Question Number 208359 by alcohol last updated on 13/Jun/24
Commented by alcohol last updated on 13/Jun/24
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Answered by Berbere last updated on 17/Jun/24
f(x)=x^n +x−1⇒f′(x)=nx^(n−1) +1≥1;f increase f(0)=−1,f(1)=1 ∃c_n ∣f(x_n )=0 x_n ^n +x−1=0 f_(n+1) (x_n )=x_n ^(n+1) +x_n −1≤x_n ^n +x_n −1=0 x_(n+1) >x_n ⇒x_n cv x_n =l if l<1⇒lim_(n→∞) l^n =0⇒l−1=0⇒l=1 absurd ⇒l=1 since 0≤x_n ≤1 1−x_n =x_n ^n =u_n ⇒u_n →0⇒ 1−x_n =x_n ^n ⇒nln(x_n )−ln(1−x_n )=0 x_n =x_n =1−u_n nln(1−u_n )−ln(u_n )=0⇒u_n solution of f(x)=nln(1−x)−ln(x) u_n ∈]0,1[ f′(x)=((−n)/(1−x))−(1/x)<0 f(((ln(n))/(2n)))=nln(1−((ln(n))/(2n)))−ln(((ln(n))/(2n))) ∼−((ln(n))/2)−ln(ln(n))+ln(2n) =((ln(n))/2)−ln(ln(n))>0 for n enough big f(2((ln(n))/n))=nln(1−((2ln(n))/n))−ln(((2ln(n))/n)) ∼−2ln(n)−ln(2ln(n)+ln(n)∼−ln(n)<0 for n enough big f(((ln(n))/(2n)))>;f(((2ln(n))/n))>0 f(u_n )=0 f decrease ⇒ ((ln(n))/(2n))<u_n <((2ln(n))/n)
f(x)=xn+x−1⇒f′(x)=nxn−1+1⩾1;fincreasef(0)=−1,f(1)=1∃cn∣f(xn)=0xnn+x−1=0fn+1(xn)=xnn+1+xn−1⩽xnn+xn−1=0xn+1>xn⇒xncvxn=lifl<1⇒limn→∞ln=0⇒l−1=0⇒l=1absurd⇒l=1since0⩽xn⩽11−xn=xnn=un⇒un→0⇒1−xn=xnn⇒nln(xn)−ln(1−xn)=0xn=xn=1−unnln(1−un)−ln(un)=0⇒unsolutionoff(x)=nln(1−x)−ln(x)un∈]0,1[f′(x)=−n1−x−1x<0f(ln(n)2n)=nln(1−ln(n)2n)−ln(ln(n)2n)∼−ln(n)2−ln(ln(n))+ln(2n)=ln(n)2−ln(ln(n))>0fornenoughbigf(2ln(n)n)=nln(1−2ln(n)n)−ln(2ln(n)n)∼−2ln(n)−ln(2ln(n)+ln(n)∼−ln(n)<0fornenoughbigf(ln(n)2n)>;f(2ln(n)n)>0f(un)=0fdecrease⇒ln(n)2n<un<2ln(n)n

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